Answer
The statement $x=\frac{-4\pm \sqrt{{{4}^{2}}-4\cdot 2\cdot 5}}{2\cdot 2}$ simplifies to $x=-1\pm i\frac{\sqrt{6}}{2}$.
Work Step by Step
Consider the expression $x=\frac{-4\pm \sqrt{{{4}^{2}}-4\cdot 2\cdot 5}}{2\cdot 2}$.
Simplify the expression.
\[\begin{align}
& x=\frac{-4\pm \sqrt{16-40}}{4} \\
& =\frac{-4\pm \sqrt{-24}}{4}
\end{align}\]
The principal square root of a negative number is such that for any positive real number $b$,
$\sqrt{-b}=i\sqrt{b}$
Write the expression in terms of an imaginary unit $i$.
\[x=\frac{-4\pm i\sqrt{24}}{4}\]
Make the factors of $24$.
\[x=\frac{-4\pm i\sqrt{4\cdot 6}}{4}\]
Use the property of product of radicals $\sqrt{ab}=\sqrt{a}\cdot \sqrt{b}$.
\[\begin{align}
& x=\frac{-4\pm i\sqrt{4}\cdot \sqrt{6}}{4} \\
& =\frac{-4\pm 2i\sqrt{6}}{4} \\
& =\frac{-4}{4}\pm \frac{2i\sqrt{6}}{4} \\
& =-1\pm i\frac{\sqrt{6}}{2}
\end{align}\]
Therefore, the correct fill for the blank is
\[-1\pm i\frac{\sqrt{6}}{2}\]
.
